Question

Solve for x: log(3x) + log(x + 4) = log(15).

Answer 1

Answer:

x=1

Step-by-step explanation:

The given equation is

$\log(3x)+\log (x+4)=\log (15)$

We need to solve the equation for x.

Using the product property of logarithm we get

$\log(3x(x+4))=\log (15)$            $[\because \log (ab)=\log a+\log b]$

$\log(3x^2+12x)=\log (15)$

On comparing both sides we get

$3x^2+12x=15$

$3x^2+12x-15=0$

Find factors using grouping method.

$3x^2+15x-3x-15=0$

$3x(x+5)-3(x+5)=0$

$(x+5)(3x-3)=0$

Using zero product property we get

$x+5=0\Rightarrow x=-5$

$3x-3=0\Rightarrow x=1$

For x=-5, 3x=-15 and log is not defined for negative values.

It means x=-5 is an extraneous solution.

Therefore, x=1 is the only valid solution of the given equation.

Answer 2

Answer:

The solutions are x = 1 and x = -5

Step-by-step explanation:

Hi there!

First, let´s write the equation:

log(3x) + log(x + 4) = log(15)

Apply logarithm property: log(3x) = log(3) + log(x)

log(3) + log(x) + log(x + 4) = log(15)

Substract log(3) from both sides of the equation

log(x) + log(x+4) = log(15) - log(3)

Apply logarithm property: log(15) - log(3) = log(15/3) = log(5)

log(x) + log(x + 4) = log(5)

Apply logarithm property: log(x) + log(x+4) = log(x (x+4)) = log(x² + 4x)

log(x² + 4x) = log(5)

Apply logarithm equality rule: if log(x² + 4x) = log(5), then x² + 4x = 5

x² + 4x = 5

Substract 5 from both sides

x² + 4x - 5 = 0

Using the quadratic formula (a = 1, b = 4, c = -5)

x = 1 and x = -5

The solutions are x = 1 and x = -5

Have a nice day!